Priority Queues Two kinds of priority queues: Min priority queue. Max priority queue. Nov 4, 20091

Min Priority Queue Collection of elements. Each element has a priority or key. Supports following operations:  isEmpty  size  add/put an element into the priority queue  get element with min priority  remove element with min priority Nov 4, 20092

Max Priority Queue Collection of elements. Each element has a priority or key. Supports following operations:  isEmpty  size  add/put an element into the priority queue  get element with max priority  remove element with max priority Nov 4, 20093

Complexity Of Operations Two good implementations are heaps and leftist trees. isEmpty, size, and get => O(1) time put and remove => O(log n) time where n is the size of the priority queue Nov 4, 20094

Applications Sorting use element key as priority put elements to be sorted into a priority queue extract elements in priority order  if a min priority queue is used, elements are extracted in ascending order of priority (or key)  if a max priority queue is used, elements are extracted in descending order of priority (or key) Nov 4, 20095

Sorting Example Sort five elements whose keys are 6, 8, 2, 4, 1 using a max priority queue.  Put the five elements into a max priority queue.  Do five remove max operations placing removed elements into the sorted array from right to left. Nov 4, 20096

After Putting Into Max Priority Queue Sorted Array Max Priority Queue Nov 4, 20097

After First Remove Max Operation Sorted Array Max Priority Queue Nov 4, 20098

After Second Remove Max Operation Sorted Array Max Priority Queue Nov 4, 20099

After Third Remove Max Operation Sorted Array Max Priority Queue Nov 4,

After Fourth Remove Max Operation Sorted Array Max Priority Queue Nov 4,

After Fifth Remove Max Operation Sorted Array Max Priority Queue Nov 4,

Complexity Of Sorting Sort n elements.  n put operations => O(n log n) time.  n remove max operations => O(n log n) time.  total time is O(n log n).  compare with O(n 2 ) for sort methods Nov 4,

Heap Sort Uses a max priority queue that is implemented as a heap. Initial put operations are replaced by a heap initialization step that takes O(n) time. Nov 4,

Machine Scheduling  m identical machines (drill press, cutter, sander, etc.)  n jobs/tasks to be performed  assign jobs to machines so that the time at which the last job completes is minimum Nov 4,

Machine Scheduling Example 3 machines and 7 jobs job times are [6, 2, 3, 5, 10, 7, 14] possible schedule A B C time > Nov 4,

Machine Scheduling Example Finish time = 21 Objective: Find schedules with minimum finish time. A B C time > Nov 4,

LPT Schedules Longest Processing Time first. Jobs are scheduled in the order 14, 10, 7, 6, 5, 3, 2 Each job is scheduled on the machine on which it finishes earliest. Nov 4,

LPT Schedule [14, 10, 7, 6, 5, 3, 2] A B C Finish time is 16! Nov 4,

LPT Schedule LPT rule does not guarantee minimum finish time schedules. Usually LPT finish time is much closer to minimum finish time. Minimum finish time scheduling is NP-hard. Nov 4,

NP-hard Problems Infamous class of problems for which no one has developed a polynomial time algorithm. That is, no algorithm whose complexity is O(n k ) for any constant k is known for any NP- hard problem. The class includes thousands of real-world problems. Highly unlikely that any NP-hard problem can be solved by a polynomial time algorithm. Nov 4,

NP-hard Problems Since even polynomial time algorithms with degree k > 3 (say) are not practical for large n, we must change our expectations of the algorithm that is used. Usually develop fast heuristics for NP-hard problems.  Algorithm that gives a solution close to best.  Runs in acceptable amount of time. LPT rule is good heuristic for minimum finish time scheduling. Nov 4,

Complexity Of LPT Scheduling Sort jobs into decreasing order of task time.  O(n log n) time (n is number of jobs) Schedule jobs in this order.  assign job to machine that becomes available first  must find minimum of m (m is number of machines) finish times  takes O(m) time using simple strategy  so need O(mn) time to schedule all n jobs. Nov 4,

Using A Min Priority Queue Min priority queue has the finish times of the m machines. Initial finish times are all 0. To schedule a job remove machine with minimum finish time from the priority queue. Update the finish time of the selected machine and put the machine back into the priority queue. Nov 4,

Using A Min Priority Queue m put operations to initialize priority queue 1 remove min and 1 put to schedule each job each put and remove min operation takes O(log m) time time to schedule is O(n log m) overall time is O(n log n + n log m) = O(n log (mn)) Nov 4,

Huffman Codes Useful in lossless compression. Nov 4,

Min Tree Definition Each tree node has a value. Value in any node is the minimum value in the subtree for which that node is the root. Equivalently, no descendent has a smaller value. Nov 4,

Min Tree Example Root has minimum element. Nov 4,

Max Tree Example Root has maximum element. Nov 4,

Min Heap Definition complete binary tree min tree Nov 4,

Min Heap With 9 Nodes Complete binary tree with 9 nodes. Nov 4,

Min Heap With 9 Nodes Complete binary tree with 9 nodes that is also a min tree Nov 4,

Max Heap With 9 Nodes Complete binary tree with 9 nodes that is also a max tree Nov 4,

Heap Height Since a heap is a complete binary tree, the height of an n node heap is log 2 (n+1). Nov 4,

A Heap Is Efficiently Represented As An Array Nov 4,

Moving Up And Down A Heap Nov 4,

Putting An Element Into A Max Heap Complete binary tree with 10 nodes Nov 4,

Putting An Element Into A Max Heap New element is Nov 4,

Putting An Element Into A Max Heap New element is Nov 4,

Putting An Element Into A Max Heap New element is Nov 4,

Putting An Element Into A Max Heap New element is Nov 4,

Putting An Element Into A Max Heap New element is Nov 4,

Putting An Element Into A Max Heap Complete binary tree with 11 nodes Nov 4,

Putting An Element Into A Max Heap New element is Nov 4,

Putting An Element Into A Max Heap New element is Nov 4,

Putting An Element Into A Max Heap New element is Nov 4,

Complexity Of Put Complexity is O(log n), where n is heap size Nov 4,

Removing The Max Element Max element is in the root Nov 4,

Removing The Max Element After max element is removed Nov 4,

Removing The Max Element Heap with 10 nodes Reinsert 8 into the heap. Nov 4,

Removing The Max Element Reinsert 8 into the heap Nov 4,

Removing The Max Element Reinsert 8 into the heap Nov 4,

Removing The Max Element Reinsert 8 into the heap Nov 4,

Removing The Max Element Max element is Nov 4,

Removing The Max Element After max element is removed Nov 4,

Removing The Max Element Heap with 9 nodes Nov 4,

Removing The Max Element Reinsert Nov 4,

Removing The Max Element Reinsert Nov 4,

Removing The Max Element Reinsert Nov 4,

Complexity Of Remove Max Element Complexity is O(log n) Nov 4,

Initializing A Max Heap input array = [-, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] Nov 4,

Initializing A Max Heap Start at rightmost array position that has a child Index is n/2. Nov 4,

Initializing A Max Heap Move to next lower array position Nov 4,

Initializing A Max Heap Nov 4,

Initializing A Max Heap Nov 4,

Initializing A Max Heap Nov 4,

Initializing A Max Heap Nov 4,

Initializing A Max Heap Nov 4,

Initializing A Max Heap Find a home for 2. Nov 4,

Initializing A Max Heap Find a home for Nov 4,

Initializing A Max Heap Done, move to next lower array position Nov 4,

Initializing A Max Heap Find home for 1. Nov 4,

11 Initializing A Max Heap Find home for 1. Nov 4,

Initializing A Max Heap Find home for 1. Nov 4,

Initializing A Max Heap Find home for 1. Nov 4,

Initializing A Max Heap Done. 1 Nov 4,

Time Complexity Height of heap = h. Number of subtrees with root at level j is <= 2 j-1. Time for each subtree is O(h-j+1). Nov 4,

Complexity Time for level j subtrees is <= 2 j-1 (h-j+1) = t(j). Total time is t(1) + t(2) + … + t(h-1) = O(n). Nov 4,

Leftist Trees Linked binary tree. Can do everything a heap can do and in the same asymptotic complexity. Can meld two leftist tree priority queues in O(log n) time. Nov 4,

Extended Binary Trees Start with any binary tree and add an external node wherever there is an empty subtree. Result is an extended binary tree. Nov 4,

A Binary Tree Nov 4,

An Extended Binary Tree number of external nodes is n+1 Nov 4,

The Function s() For any node x in an extended binary tree, let s(x) be the length of a shortest path from x to an external node in the subtree rooted at x. Nov 4,

s() Values Example Nov 4,

s() Values Example Nov 4,

Properties Of s() If x is an external node, then s(x) = 0. Otherwise, s(x) = min {s(leftChild(x)), s(rightChild(x))} + 1 Nov 4,

Height Biased Leftist Trees A binary tree is a (height biased) leftist tree iff for every internal node x, s(leftChild(x)) >= s(rightChild(x)) Nov 4,

A Leftist Tree Nov 4,

Leftist Trees--Property 1 In a leftist tree, the rightmost path is a shortest root to external node path and the length of this path is s(root). Nov 4,

A Leftist Tree Length of rightmost path is 2. Nov 4,

Leftist Trees—Property 2 The number of internal nodes is at least 2 s(root) - 1 Because levels 1 through s(root) have no external nodes. So, s(root) <= log(n+1) Nov 4,

A Leftist Tree Levels 1 and 2 have no external nodes. Nov 4,

Leftist Trees—Property 3 Length of rightmost path is O(log n), where n is the number of nodes in a leftist tree. Follows from Properties 1 and 2. Nov 4,

Leftist Trees As Priority Queues Min leftist tree … leftist tree that is a min tree. Used as a min priority queue. Max leftist tree … leftist tree that is a max tree. Used as a max priority queue. Nov 4,

A Min Leftist Tree Nov 4,

Some Min Leftist Tree Operations put() remove() meld() initialize() put() and remove() use meld(). Nov 4,

Put Operation put(7) Nov 4,

Put Operation put(7) Create a single node min leftist tree. 7 Nov 4,

Put Operation put(7) Create a single node min leftist tree. Meld the two min leftist trees. 7 Nov 4,

Remove Min Nov 4,

Remove Min Remove the root. Nov 4,

Remove Min Remove the root. Meld the two subtrees. Nov 4,

Meld Two Min Leftist Trees Traverse only the rightmost paths so as to get logarithmic performance. Nov 4,

Meld Two Min Leftist Trees Meld right subtree of tree with smaller root and all of other tree. Nov 4,

Meld Two Min Leftist Trees Meld right subtree of tree with smaller root and all of other tree. Nov 4,

Meld Two Min Leftist Trees Meld right subtree of tree with smaller root and all of other tree. Nov 4,

Meld Two Min Leftist Trees 8 6 Meld right subtree of tree with smaller root and all of other tree. Right subtree of 6 is empty. So, result of melding right subtree of tree with smaller root and other tree is the other tree. Nov 4,

Meld Two Min Leftist Trees Swap left and right subtree if s(left) < s(right). Make melded subtree right subtree of smaller root Nov 4,

Meld Two Min Leftist Trees Make melded subtree right subtree of smaller root. Swap left and right subtree if s(left) < s(right). Nov 4,

Meld Two Min Leftist Trees Swap left and right subtree if s(left) < s(right). Make melded subtree right subtree of smaller root Nov 4,

Meld Two Min Leftist Trees Nov 4,

Initializing In O(n) Time create n single node min leftist trees and place them in a FIFO queue repeatedly remove two min leftist trees from the FIFO queue, meld them, and put the resulting min leftist tree into the FIFO queue the process terminates when only 1 min leftist tree remains in the FIFO queue analysis is the same as for heap initialization Nov 4,